Kul-24.3300 Ship Buoyancy and Stability Model Answer 13.10.2012Assignment 3.Markus Tompuri 1(4)

1. Let us consider metacentric height of the buoy:

! 00GM The buoy is stable and floats upright 00GM The buoy is unstable and floats inclined

Metacentric height of the buoy can be determined with following equation:

220000HITKGMBKBGM T

Where T is the draft of the buoy, IT is the waterplane moment of inertia and H is theheight of the buoy. Lifting force of the buoyancy is equal to the weight of the buoy:

gmg U '

Where m is the mass of the buoy and U is the water density. 33 m0.020kg1000

kg20 m

mU

Volume of displacement of the buoy is:

TD 24S

Where D is the diameter of the bottom. Thus the draft is:

m0.126m0.45m0.02044

2

3

2

SSDT

Waterplane area of the buoy is circular, so the moment of inertia can be calculatedwith following equation:

444 m0.00264

m0.4564

SSDIT

When we calculate the GM, we get:

m0.1622

m0.65m0.020m0.002

2m0.126

3

4

0 GM

00GM Buoy is unstable and floats inclined.

Kul-24.3300 Ship Buoyancy and Stability Model Answer 13.10.2012Assignment 3.Markus Tompuri 2(4)

2. If the ship has N number of tanks, the combined free surface effect to themetacenric height can be calculated with following equation:

jT

N

jjtcorr iGMGM ,

1,0

1

UU

Lets first calcualte the uncorrected metacentric height:

mmmKGKMGM 0.10.60.700

We look at three different cases based on the figure below. Length of the tanks isalways lj = 20 m. All the tanks contain ballast water, therefore,

jmkg

jt 31025, UU

Volume of the displacement is:

33

9.78041025

1080002

mkg

mkg

Lets calculate the corrected (effective) metacentric heights for different cases:

a)

mm

mmm

blGM

iGMGMN

jjTcorr

125.09.780412

16200.1

121

1

3

3

31

0

1,0

Kul-24.3300 Ship Buoyancy and Stability Model Answer 13.10.2012Assignment 3.Markus Tompuri 3(4)

b)

mm

mmm

blGM

iGMGMN

jjTcorr

781.09.780412

82020.1

1221

1

3

3

32

0

1,0

c)

mm

mmm

blGM

iGMGMN

jjTcorr

945.09.780412

42040.1

1241

1

3

3

33

0

1,0

3. Ship speed is Vs = 21 kn = 10.8 m/s

Yaw motion angular velocity:

rV

DV s

s

s 2

\

Centrifugal force acting on the center of gravity of the ship:

ss

kp VDF \U\U 2

2

Kul-24.3300 Ship Buoyancy and Stability Model Answer 13.10.2012Assignment 3.Markus Tompuri 4(4)

Having neglected the rudder forces, for the equilibrium the hydrodynamic reactionand the centrifugal force have the same magnitude.

The heeling moment is:

2TKGVM sext \U

Righting static moment (approximation of initial stability):

MU sin0GMgM st

In the equilibrium situation (steady turn) sum of the moments is zero, therefore:

q

8.6

4.081.92

82.6/8.10

550/8.10arcsin2arcsin

2sin

0sin2

0

30

0

0

I

\I

\I

MU\U

m

mmsm

msm

GMg

TKGV

GMg

TKGV

GMgTKGV

MM

sms

s

s

stext